91Ó°ÊÓ

The voltage, ${\mathrm{Î\mu }}_1=10.0\text{V}$

The voltage, ${\mathrm{Î\mu }}_2=0.500{\mathrm{Î\mu }}_1$

The resistors, $R_1=R_2=R_3=4.00\mathrm{Î\copyright }$

Apply the Kirchhoff’s loop rule and thejunction rule to each loop and solve the equations to get the current in resistance 2 and resistance 3.

Formula:

The total current is define by,

$i_3=i_1+i_2$

Apply the loop rule to the left side closed loop,

${\mathrm{Î\mu }}_1−i_1{R}_1−i_3{R}_3=0$ ….. (1)

But according to the junction rule,

$i_3=i_1+i_2$ ….. (2)

Substitute the above equation into equation (1).

$\begin{array}{l}{\mathrm{Î\mu }}_1−i_1{R}_1−(i_1+i_2)R_3=0\\ {\mathrm{Î\mu }}_1−i_1{R}_1−i_1{R}_3−i_2{R}_3=0\\ {\mathrm{Î\mu }}_1−i_1(R_1+R_3)−i_2{R}_3=0\end{array}$

Substitute $10\text{V}$ for ${\mathrm{Î\mu }}_1$, $4.00\mathrm{Î\copyright }$for $R_1$and $R_3$, you get

role="math" localid="1662701819687" $$\begin{gathered} \begin{array}{l}10\text{V}−i_1(4.00\mathrm{Î\copyright }+4.00\mathrm{Î\copyright })−i_2\left(4.00\mathrm{Î\copyright }\right)=0\\ 10\text{V}−i_1\left(8.00\mathrm{Î\copyright }\right)−i_2\left(4.00\mathrm{Î\copyright }\right)=0\\ 10\text{V}=i_1\left(8.00\mathrm{Î\copyright }\right)+i_2\left(4.00\mathrm{Î\copyright }\right)\end{array} \\ \text{5V}=i_1\left(4.00\mathrm{Î\copyright }\right)+i_2\left(2.00\mathrm{Î\copyright }\right) \end{gathered}$$ ….. (3)

Now, applying the loop rule to the right-side closed loop,

${\mathrm{Î\mu }}_2−i_2{R}_2−i_3{R}_3=0$ ….. (4)

But according to the junction rule,

$i_3=i_1+i_2$

Substitute the above equation into equation (4).

$\begin{array}{l}{\mathrm{Î\mu }}_2−i_2{R}_2−(i_1+i_2)R_3=0\\ {\mathrm{Î\mu }}_2−i_2{R}_2−i_1{R}_3−i_2{R}_3=0\\ {\mathrm{Î\mu }}_2−i_2(R_2+R_3)−i_1{R}_3=0\end{array}$

As $5\text{V}$for ${\mathrm{Î\mu }}_2$, $4.00\mathrm{Î\copyright }$for $R_2$and $R_3$, and youhave

$$\begin{gathered} 5V−i_2(4.00\mathrm{Î\copyright }+4.00\mathrm{Î\copyright })−i_1\left(4.00\mathrm{Î\copyright }\right)=0 \\ 5V=i_1\left(4.00\mathrm{Î\copyright }\right)+i_2\left(8.00\mathrm{Î\copyright }\right) \end{gathered}$$ ….. (5)

Subtract equation(3)from equation(5),

$\begin{array}{l}5\text{V}−5\text{V}=i_1\left(4.00\mathrm{Î\copyright }\right)+i_2\left(8.00\mathrm{Î\copyright }\right)−i_1\left(4.00\mathrm{Î\copyright }\right)−i_2\left(2.00\mathrm{Î\copyright }\right)\\ 0\text{V}=i_2\left(6.00\mathrm{Î\copyright }\right)\\ i_2=0\text{A}\end{array}$

Hence, the current in resistance 2 is$0\text{A}$ .

As you have equation(3),

$\text{5V}=i_1\left(4.00\mathrm{Î\copyright }\right)+i_2\left(2.00\mathrm{Î\copyright }\right)$

Substitute $0\text{A}$for $i_2$ in the above equation.

$\begin{array}{c}\text{5V}=i_1\left(4.00\mathrm{Î\copyright }\right)+\left(0\text{A}\right)\left(2.00\mathrm{Î\copyright }\right)\\ =i_1\left(4.00\mathrm{Î\copyright }\right)\end{array}$

$\begin{array}{c}{i}_1=\frac{5\text{V}}{4.00\mathrm{Î\copyright }}\\ =1.25\text{A}\end{array}$

Rewrite equatopn (2) and substitute the known values.

$\begin{array}{c}{i}_3=i_1+i_2\\ =1.25\text{A}+0\text{A}\\ =1.25\text{A}\end{array}$

Hence, the current in resistance 3 is $1.25\text{A}$.